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Jessica S.
Student at Johns Hopkins University
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SAT
TutorMe
Question:

How do I time manage on the SAT?

Jessica S.
Answer:

Preparation beforehand is key. When I took the SATs, it was hard for me to adapt to the style and format of the question initially. As such, taking multiple practice exams helped me understand the format more, which resulted in spending a lot less time focusing on understanding the question and actually solving it. Additionally, it is always better to go through the exam multiple times. Do not stop and spend ten minutes on a question that you do not get. If you don't get it immediately or do not think you will be able to get the answer within a minute, move onto the next question. It is better to spend your time on multiple questions that you have a vague recollection of how to do than focus all of your energy on one very difficult and confusing problem. Also, sometimes, answering other questions will help you remember how to answer another question. Again, practice helps, so do as many practice questions as you can. Some question formats are very similar--it's a matter of exposure.

College Admissions
TutorMe
Question:

How can I stand out from other applicants?

Jessica S.
Answer:

Depending on what your stats are and what college you're applying for, there are many things you can do to stand out. If, for example, all of your stats match the school's typical student, a more risky essay could potentially help you stand out. Granted, this should be a calculated risk, but if you are passionate in your essay, there is a good chance that you will stand out. Trying to think of unconventional ways to show the school what they are looking for can help, but keep in mind that schools care mostly about seeing what YOU are passionate about, what you strive to do, what you can accomplish at the school. At the end of the day, it's hard to predict what the school will decide. But, with enough guidance understanding what the school is looking for, you can increase your chances of being seen as a strong applicant, and thus increase your chance of being accepted.

Algebra
TutorMe
Question:

Solve a system of equations using either elimination or substitution. 5x - 3y = 24 -y = 6x - 10

Jessica S.
Answer:

1. Regardless of whether you want to use elimination or substitution to solve this problem, when you see a systems of equation problem, the first thing that you should do is rearrange it so that the equations are in the same format. For this problem, move the 6x of the second equation to the left side. Don't forget to change the 6x to a -6x: 5x - 3y = 24 -6x - y = -10 2a. Elimination method: i. If you are using ELIMINATION, your goal is to find a way to make the sum of either the x's or the y'x in the 2 equations equal to 0 so that you can successfully eliminate one of the variables. Looking at the 2 variables and their respective values (5x and -6x, and -3y and -y), the simpler method would be to change the -y into a +3y (so that when you add the 2 equations together, the -3y and +3y cancel each other out). To do this, we multiple the entire 2nd equation by -3: 5x - 3y = 24 --> 5x - 3y = 24 (-6x -y = -10)-3 --> 18x + 3y = 30 ii. Now, we add the 2 equations together: 5x - 3y = 24 + 18x + 3y = 30 ---------------------------- 23x + 0y = 54 = 23x = 54 iii. Simplify the equation to find the value of x by dividing both sides by 23: 23x = 54 23x/23 = 54/23 x = 2.348 iv. Plug the value of x back into either of the 2 equations to find the value of y. -6x - y = -10 -6(2.348) - y = -10 -14.088 - y = -10 Isolate y ( -14.088 - y) +14.088 = -10 +14.088 -y = 4.088 y = -4.088 2b. Substitution Method i. If you are using the SUBSTITUTION method, you want to isolate one of the variables on one side of the equation and plug in the expression on the other side into the other equation. For this problem, it is easier to isolate the y of the 2nd equation: 5x - 3y = 24 -6x - y = -10 <-- isolate the y by moving the y to the right side by adding +y to both sides (-6x - y) + y = -10 + y -6x = -10 + y <-- Move the -10 to the left side by adding +10 to both sides -6x + 10 = -10 +y +10 -6x + 10 = y <-- The left side of the equation becomes the expression that you want to plug into the other equation. ii. Now that you have isolated a variable, replace the expression into the respective variable (in this case y) of the other equation. Given y = -6x + 10 and 5x - 3y = 24, 5x - 3(-6x + 10) = 24 ii. Simplify and isolate x to find the value of x. 5x + 18x - 30 = 24 23 x = 54 23x/23 = 54/23 x = 2.348 iii. Plug in the value of x into one of the equations to find the value of y: y = -6x + 10 y = -6(2.348) + 10 y = -4.088

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